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$y= 0.4\ x^2 \cdots (Given)\ Hence,\ \dfrac {dy} {dx}= 0.8x\ and \dfrac {d^2y} {dx^2}=0.8\ At\ x=2, \dfrac {dy} {dx}= 0.8 \times 2= 1.6\ and \dfrac {d^2y} {dx^2}=0.8\$
Radius of curvature (ρ)=[1+ (dy/dx)2]3/2d2y/dx2[1+(1.6)2]3/20.8=8.3962 m.
x=2, velocity(v)=6m/s and tangential acceleration (at)=3m/s2⋯(Given)With the horizontal, let velocity vector make an angle θHence, tanθ=m=slope of tangent=dydx=1.6
θ=58 ∘Velocity along x−direction (vx)=vcosθ=6.cos58=3.1795 m/sVelocity along y−direction (vy)=vsinθ=6.sin58=5.0883 m/s
Normal acceleration (an)=v2ρ=628.3962=4.2877 m/s2a2=a12+a22=32+4.28772=27.3844Hence, (a)=5.2330 m/s2.
Let acceleration vector make an angle ′α′ with at.Hence, α=tan−1anat=tan−14.28773=55.0204 ∘With horizontal, angle made by acceleration =θ+α=55.0204+58=113.0204∘
Hence,Velocity along x−direction (vx)=3.1795 m/s; Velocity along y−direction; (vy)=5.0883 m/s and acceleration=5.2330 m/s2,∠=113.0204 ∘